Law of Sines/Proof 1
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Theorem
Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.
Then:
- $\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$
where $R$ is the circumradius of $\triangle ABC$.
Proof
Construct the altitude from $B$.
From the definition of sine:
- $\sin A = \dfrac h c$ and $\sin C = \dfrac h a$
Thus:
- $h = c \sin A$
and:
- $h = a \sin C$
This gives:
- $c \sin A = a \sin C$
So:
- $\dfrac a {\sin A} = \dfrac c {\sin C}$
Similarly, constructing the altitude from $A$ gives:
- $\dfrac b {\sin B} = \dfrac c {\sin C}$
$\blacksquare$